Question

A car starts from rest, attains a velocity of $36km{h}^{-1}$ with an acceleration of $0.2m{s}^{-2}$, travels $9km$ with this uniform velocity and then comes to halt with a uniform deceleration of $0.1m{s}^{-2}$. The total time of travel of the car is

• A. $1050s$
• B. $1000s$
• C. $950s$
• D. $900s$

Answer 1

Answer: (A)

Let the car be accelerated from $A$ to $B$. It moves with uniform velocity from $B$ to $C$ and then moves with uniform deceleration from $C$ to $D$.
For the motion of car from $A$ to $B$,
$u=0$, $v=36km{h}^{-1}=36\times \frac{5}{18}m{s}^{-1}=10m{s}^{-1}$, $a=0.2m{s}^{-2}$
Time taken, $t_1=\frac{v-u}{a}=\frac{10m{s}^{-1}-0}{0.2m{s}^{-2}}=50s$
For the motion of car from $B$ to $C$
$S=9km=9000m$
Time taken, $t_2=\frac{9000m}{10m{s}^{-1}}=900s$
For the motion of car from $C$ to $D$,
$v=0$, $u=10m{s}^{-1}$, $a=-0.1m{s}^{-2}$
Time taken, $t_3=\frac{0-10m{s}^{-1}}{-0.1m{s}^{-2}}=100s$
Total time taken $=t_1+t_2+t_3$ ;
$=50s+900s+100s=1050s$